Doesn't the number of explored nodes with IDA* increase linearly?
No misunderstanding. IDA* is indeed n^2 vs n for A* in certain situations.
See
https://pdfs.semanticscholar.org/388c/0a934934a9e60da1c22c050566dbcd995702.pdf
for a reference.
IDA* only works in certain (not uncommon) situations. If there are many states with the same costs, and this number grows exponentially with the cost, then IDA* will be O(n). An example of such a problem is the 8 sliding piece puzzle.
IDA* will however, as noted, become exponential when all costs differ. Examples include continuous costs, or discrete values spanning a large range.
If you really wanted to apply IDA* to e.g. a map, a workable approach would likely be to round all values - or perform a log+floor operation.